Necessary and Sufficient Conditions for a CDF This is an attempt to prove Theorem 1.5.3. in Casella and Berger. Note that the only things that have been proven are really basic set-theory with $\mathbb{P}$ (a probability measure) theorems (e.g., addition rule). Recall for a random variable $X$, we define $$F_X(x) = \mathbb{P}(X \leq x)\text{.}$$

Theorem. $F$ is a CDF iff:
  
  
*
  
*$\lim\limits_{x \to -\infty}F(x) = 0$
  
*$\lim\limits_{x \to +\infty}F(x) = 1$
  
*$F$ is nondecreasing.
  
*For all $x_0 \in \mathbb{R}$, $\lim\limits_{x \to x_0^{+}} F(x)= F(x_0)$
  

$\Longrightarrow$ If $F$ is a CDF of $X$, by definition,
$$F_{X}(x) = \mathbb{P}(X \leq x) = \mathbb{P}\left(\{s_j \in S: X(s_j) \leq x\} \right) $$
where $S$ denotes the overall sample space. 
$(3)$ is easy to show. Suppose $x_1 \leq x_2$. Then notice
$$\{s_j \in S: X(s_j) \leq x_1\} \subset \{s_j \in S: X(s_j) \leq x_2\}$$
and therefore by a Theorem,
$$\mathbb{P}\left(\{s_j \in S: X(s_j) \leq x_1\}\right) \leq \mathbb{P}\left(\{s_j \in S: X(s_j) \leq x_2\}\right)$$
giving $F_{X}(x_1) \leq F_{X}(x_2)$, hence $F$ is nondecreasing.
I suppose $(1)$ and $(2)$ aren't consequences of anything more than saying that $\{s_j \in S: X(s_j) \leq -\infty\} = \varnothing$ and $\{s_j \in S: X(s_j) \leq +\infty\} = S$ (unless I'm completely wrong here). But this seems to suggest that $$\lim_{x \to -\infty}\mathbb{P}(\text{blah}(x)) = \mathbb{P}(\lim_{x \to -\infty}\text{blah}(x))$$
where $\text{blah}(x)$ is a set dependent on $x$. At this point of the text, this hasn't been proven (if it's even true).
I'm not sure how to show $(4)$.
$\Longleftarrow$ I don't know how to prove sufficiency. Casella and Berger state that this is "much harder" than necessity, and we have to establish that there is a sample space, a probability measure, and a random variable defined on the sample space such that $F$ is the CDF of this random variable, but this isn't enough detail for me to go on.
 A: This is from Probability: Theory and Examples by Rick Durrett, Theorem 1.2.1 and Theorem 1.2.2.
Theorem 1.2.1. Any distribution function $F$ has the properties that you listed.
Proof. 


*

*(Non-decreasing) If $x\leq y$ then $\{X\leq x\}\subset\{X\leq y\}$, and the monotonicity of Lebesgue measure gives $P(X\leq x)\leq P(X\leq y)$.

*(Limitations) It is because $\lim_{x\to \infty} \{X\leq x\}=\Omega$, $\lim_{x\to -\infty} \{X\leq x\}$.

*(Right continuous) It is because $\lim_{y\to x^+} \{X\leq y\}=\{X\leq x\}$.
Theorem 1.2.2. If $F$ satisfies the condition that you listed, it is the distribution function of some random variable.
Proof. Let $\Omega=(0,1),\mathcal{F}=$ the Borel sets and $P=$ Lebesgue measure. If $\omega\in(0,1)$, let $X(\omega)=\sup\{y:F(y)<\omega\}$.
If we can prove $\{\omega:X(\omega)\leq x\}=\{\omega:\omega\leq F(x)\}$, the desired result follows from since $P(\omega:\omega\leq F(x))=F(x)$.
To check $\{\omega:X(\omega)\leq x\}=\{\omega:\omega\leq F(x)\}$, we observe that if $\omega\leq F(x)$ then $X(\omega)\leq x$, since $x\notin \{y:F(y)<\omega\}$. On the other hand, if $\omega>F(x)$, then since $F$ is right continuous, there is an $\epsilon>0$ so that $F(x+\epsilon)<\omega$ and $X(\omega)\geq x+\epsilon>x$.
